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Iterative schemes for computing fixed points of nonexpansive mappings in Banach spaces
Jean-Philippe Chancelier
To cite this version:
Jean-Philippe Chancelier. Iterative schemes for computing fixed points of nonexpansive mappings in
Banach spaces. 2007. �hal-00184662v3�
hal-00184662, version 3 - 7 Dec 2007
Iterative schemes for computing fixed points of nonexpansive mappings in Banach spaces
Jean-Philippe Chancelier
∗December 7, 2007
Abstract
LetX be a real Banach space with a normalized duality mapping uni- formly norm-to-weak⋆ continuous on bounded sets or a reflexive Banach space which admits a weakly continuous duality mappingJΦwith gaugeφ.
Letfbe anα-contractionand{Tn}a sequence of nonexpansive mapping, we study the strong convergence of explicit iterative schemes
xn+1=αnf(xn) + (1−αn)Tnxn (1) with a general theorem and then recover and improve some specific cases studied in the literature [17, 8, 13, 14, 3, 9].
1 Introduction and preliminaries
LetX be a real Banach space,Ca nonempty closed convex subset ofX. Recall that a mappingT :C7→C isnonexpansiveif kT(x)−T(y)k ≤ kx−yk for all x,y∈C and a mappingf :C7→Cis an α-contractionif there existsα∈(0,1) such thatkf(x)−f(y)k ≤αkx−yk for allx,y∈C.
We denote byF ix(T) the set of fixed points ofT, that is
F ix(T)def={x∈C : T x=x} (2) and ΠC will denote the collection of contractions onC.
LetX be a real Banach space. The (normalized) duality map J:X7→X⋆, whereX⋆is the dual space ofX, is defined by :
J(x)def=n
x⋆∈X⋆ : hx, x⋆i=kxk2=kx⋆k2o and there holds the inequality
kx+yk2≤ kxk2+ 2hy, j(x+y)i wherex, y ∈X andj(x+y)∈J(x+y).
∗Cermics, ´Ecole Nationale des Ponts et Chauss´ees, 6 et 8 avenue Blaise Pascal, 77455, Marne la Vall´ee, Cedex, France
Recall that ifCandF are nonempty subsets of a Banach spaceX such that C is nonempty closed convex and F ⊂ C, then a map R : C 7→F is called a retractionfromC ontoF ifR(x) =xfor allx∈F. A retractionR:C7→F is sunnyprovided R(x+t(x−R(x))) = R(x) for allx∈ C and t≥0 whenever x+t(x−R(x))∈ C. Asunny nonexpansive retraction is a sunny retraction, which is also nonexpansive.
Suppose thatF is the non empty fixed point set of a nonexpansive mapping T : C 7→ C, that is F = F ix T 6= ∅ and assume that F is closed. For a givenu∈C and everyt ∈(0,1) there exists a fixed point, denoted xt, of the (1−t)-contraction tu+ (1−t)T. Then we define Q : C 7→ F = F ix(T) by Q(u)def=σ- limt→0xt when this limit exists (σ- lim denotes the strong limit). Q will also be denoted byQF ix(T)when necessary and note that it is easy to check that, when it exists,Qis a nonexpansive retraction.
Consider nowf anα-contraction, thenQF ix(T)◦f is also an α-contraction and admits therefore a unique fixed point ˜x=QT◦f(˜x). We define byQ(f) or QF ix(T)(f) the mappingQ(f) : ΠC →F ix(T) such that :
Q(f)def= ˜x where x˜= (QF ix(T)◦f)(˜x). (3) Fort∈(0,1) we can also find a fixed point, denotedxft of the (1−(1−t)α)- contractiontf+(1−t)Tand if limt→0xft is well defined we can define a mapping
e
Q: ΠC7→F ix(T) by : Q(fe )def= lim
t→0xft where xft =tf(xft) + (1−t)T xft (4) We then gather know theorems under which Q, Q and Qe are defined and give relations between them.
WhenX is a uniformly smooth Banach space, denoted by Bus, It is known [17, Theorem 4.1] thatQe(f) is well defined and equal toQ(f) and ˜x=Q(f) is characterized by :
h˜x−f(˜x), J(˜x−p)i ≤0 for allp∈F =F ix(T). (5) A special case is whenf is a constant function u(x) =u. Then [17, Theorem 4.1] shows thatQis well defined and that Q(u) =Q(u) =PF ix Tu(where PS
is the metric projection onS). IfX is a smooth Banach space,R:C7→F is a sunny nonexpansive retraction [6] if and only if the following inequality holds : hx−Rx, J(y−Rx)i ≤0 for allx∈C andy∈F. (6) Q is thus the unique sunny non expansive retraction from C to F ix T. [17, Theorem 4.1] was already known in the case f constant and in the context of Hilbert spaces [17, Theorem 3.1] and [11, Theorem 2.1].
The same existence and characterization results can be found firstly whenX is a reflexive Banach space which admits a weakly continuous duality mapping JΦwith gaugeφ, denoted byBrwsc, in [18, Theorem 3.1] (with f constant) and
[14, Theorem 2.2] (whereJ is the (normalized) duality mapping). Note that the limitation of f constant in [18] can be relaxed with [15]. Secondly when X is a reflexive and a strictly convex Banach space with a uniformly Gˆateaux differentiable norm, denoted byBrug, [13, Theorem 3.1]. Note that in this three Banach spaces cases listed here the normalized duality mapping is shown to be single valued.
The aim of this paper is to study the strong convergence of iterative schemes : xn+1=αnf(xn) + (1−αn)Tnxn (7) whenX can be a Bus, or aBrwsc, or a Brug real Banach space and {Tn} is a sequence of nonexpansive mappings which share at least a common fixed point.
We give a general framework to show that {xn} will converge strongly to ˜x where ˜xis the unique solution of (5) for a fixed nonexpansive mappingTrelated to the sequence {Tn}. The key ingredient is the fact that Lemma 26 given in section 3 is valid in the three previous context. Then we show that by specifying the sequenceTn we can recover and extend some known convergence theorems [17, 8, 13, 14, 3, 9]. Note also that in equation (7), f is an α-contraction, but following [15] it is easy to show thatf can be replaced by a Meir-Keeler contraction (Lemma 31 in section 3 is devoted to this extension). The paper is organized as follows : a main theorem is proved in section 3 using a set of lemmas which are postponed to the last section of the paper and which are verbatim or slight extensions of know results. Then in a collection of subsections, known convergence theorems are revisited with shorter proofs.
2 Main theorem
In the sequel aB real Banach space, will denote when not specifically stated a real Banach space with a normalized duality mapping uniformly norm-to-weak⋆ continuous on bounded sets (which is the case for Bus or Brug) or a reflexive Banach space which admits a weakly continuous duality mappingJΦwith gauge φ(Brwsc).
H1,N: For a fixed given N ≥ 1 and a given sequence{αn}, a sequence of mappings {Tn} will be said to verify H1,N, if for a given bounded sequence {zn}, we have
k(1−αn+N)Tn+Nzn−(1−αn)Tnznk ≤δnM (8) with either (i)P∞
0 |δn|<∞or (i′) lim supn→∞δn/αn≤0 andM a constant.
Remark 1 Note that using Lemma 30{δn}can be replaced by{µn+ρn}where {µn} satisfies(i)and{ρn} satisfies(i′).
Remark 2 Note that when αn ∈(0,1)we have :
k(1−αn+N)Tn+Nzn−(1−αn)Tnznk ≤ |αn+N−αn|kTn+Nznk+kTn+Nzn−Tnznk.
(9)
Thus, when {αn} satisfies H3,N (given below), if for each bounded sequence {zn},{Tnzn} is bounded and either(vi) P∞
n=0kTn+Nzn−Tnznk<∞ or(vi′) kTn+Nzn−Tnznk/αn→0 thenH1,N is satisfied (again using previous remark about mixing between conditions with or without prime). In the previous case, H1,N is thus implied by H′1,N which is stated now :
H′1,N: For a fixed givenN ≥1 and a given sequence{αn} which satisfies H3,N a sequence of mappings{Tn}will be said to verifyH′1,N, if given bounded sequence{zn}, we havekTn+Nzn−Tnznk ≤ρn with either (vi)P∞
n=0ρn<∞ or (vi′)ρn/αn→0.
H2,p: For a givenp∈X, a sequence{xn} will be said to verifyH2,pif we have
lim sup
n→∞
hf(p)−p, J(xn−p)i ≤0. (10) H3,N: For a fixed given N ≥ 1, a sequence of real numbers {αn} will be said to verify H3,N if the sequence {αn} is such that (i) αn ∈ (0,1), (ii) limn→∞αn = 0, (iii)P∞
n=0αn=∞and either (iv)P∞
n=0|αn+N−αn|<∞or (iv′) limn→∞(αn+N/αn) = 1.
We can now formulate the main theorem of the paper :
Theorem 3 Let X be a B real Banach space,C a closed convex subset of X, Tn : C 7→C a sequence of nonexpansive mapping, T a nonexpansive mapping and f ∈ ΠC. We assume that F ix(T)6= ∅ and that for all n ∈ N F ix(T) ⊂ F ix(Tn). Let{αn} be a sequence of real numbers for which there exists a fixed N ≥1such thatH3,N is satisfied and suppose that there existsp∈F ix(T)such thatH2,pis satisfied, then the sequence{xn}defined by (34) converges strongly top.
Proof : The proof uses a set of Lemmas which are given in section 3. Sincep is inF ix(Tn) for allnwe can use Lemma 23 to obtain the boundedness of the sequence{xn}. Thus we can conclude using Lemma 28.
Corollary 4 Assume that the hypothesis of Theorem 3 except H2,p are satis- fied. Suppose thatH1,N orH′1,N is satisfied and that for each bounded sequence {yn}, the sequence kTnyn−T ynk →0. Then the conclusion of Theorem 3 re- mains forp=Q(f).
Proof : We just need to prove that H2,pis satisfied for p=Q(f). We first show that ifH′1,Nis satisfied thenH1,Nis also satisfied. As in previous theorem {xn}is a bounded sequence. Then, letp∈F ix(T), we have :
kTnxn−T xnk ≤ kTnxn−Tnpk+kTnp−T pk+kT p−T xnk
≤ 2kxn−pk+kTnp−T pk.
and since kTnp−T pk → 0 by hypothesis we have that {Tn(xn)} is bounded.
As shown in remark 2 we are within the case whereH1,N is implied byH′1,N.
Applying Lemma 24 and Corollary 25 we obtain the convergence ofkT xn−xnk.
We can then apply Lemma 26 to obtainH2,pforp=Q(f).
Corollary 4 can be extended as follows when a constantT cannot be found.
Corollary 5 Assume that the hypothesis of Theorem 3 except H2,p are satis- fied. Suppose thatH′1,N is satisfied and that {Tnxn} is bounded and that from each subsequenceσ(n)we can extract a subsequenceµ(n)and find a fixed map- pingTµ such that
kTµ(n)xµ(n)−Tµxµ(n)k →0.
IfF =F ix(Tµ)does not depend onµ, then the conclusion of Theorem 3 remains forp=QF(f).
Proof : We just need to prove that H2,p is satisfied for p =Q(f). Using remark 2 we are in the case whereH1,N is implied by H′1,N. Using H1,N we first easily obtain thatkxn−Tnxnk →0 by an argument similar to Corollary 25. ThenH2,pforp=Q(f) follows from Corollary 27.
We can now consider the case of composition. Assume that{Tn1} and{Tn2} satisfyH′1,N with sequences denoted by ρin. Assume also that for a bounded sequence {zn} then the sequences {Tn+N2 zn} and {Tn+N1 Tn+N2 zn} and also bounded. Then it is straightforward, since the mappingsTn1 are nonexpansive, that :
kTn+N1 Tn+N2 zn−Tn1· · ·Tn2znk ≤ ρ1n+kTn+N2 zn−Tn2znk. Thus the compositionTn1◦Tn2 satisfyH′1,N withρn
def=ρ1n+ρ2n. This lead us to propose the following Corollary for dealing with composition :
Corollary 6 Assume that the hypothesis of corollary 5 are satisfied for the sequence {Tn1} with H′1,N and for {Tn2} also with H′1,N. Then the conclusion of Theorem 3 remains for the sequence {Tn1◦Tn2} with p = QF(f) and F = F ix(Tµ1◦Tρ2).
Proof :As pointed out before the statement of the corollary the composi- tion Tn1◦Tn2 satisfy H′1,N. Consider a subsequence σ(n) we can find first a subsequenceµ2(n) andµ2such that :
kTµ(n)2 xµ(n)−Tµ2xµ(n)k →0.
Then, using properties of theTn1sequence, we can re-extract a new subsequence ρ(n) andρsuch that :
kTρ(n)1 Tρ(n)2 xρ(n)−Tρ1Tρ(n)2 xρ(n)k →0.
Since we have :
kTρ(n)1 Tρ(n)2 xρ(n)−Tρ1Tµ2xρ(n)k ≤ kTρ(n)1 Tρ(n)2 xρ(n)−Tρ1Tρ(n)2 xρ(n)k + kTρ(n)2 xρ(n)−Tρ2xρ(n)k
When obtain the conclusion for the composition.
Recall that a mapping T is attracting non expansive if it is nonexpansive and satisfies :
kT x−pk<kx−pkfor allx6∈F ix T andp∈F ix T. (11) In particular afirmly nonexpansivemapping,i.ekT x−T yk2≤ hx−y, T x−T yi is attracting nonexpansive [6].
Remark 7 In the previous corollary, we obtain a fixed point of a composition and in practice the aim is to obtain a common fixed point of two mappings. If the mappings Tµ1 and Tρ2 are attracting, have a common fixed point and Tµ1 or Tρ2is attracting then we will haveF ix Tµ1∩F ix Tρ2=F ix Tµ1◦Tρ2. The proof is contained in [1, Proposition 2.10 (i)] and given in Lemma 32 for completeness.
Remark 8 Note that if X is a strictly convex Banach space, then for λ ∈ (0,1) the mapping Tλ
def= (1−λ)I+λT is attracting nonexpansive when T is nonexpansive. Extension to a set of N operators is immediate by induction.
This gives a way to build attracting nonexpansive mappings and mixed with previous remark it gives [16, Proposition 3.1].
Remark 9 Note also that, when X is strictly convex, an other way to obtain F =∩iF ix(Ti) for a sequence of nonexpansive mappings {Ti} is to use T = P
iλiTi with a sequence {λi} of real positive numbers such that P
iλi = 1 [2, Lemma 3].
2.1 Example 1
Theorem 10 [17, Theorem 4.2] Let X be a B real Banach space, C a closed convex subset ofX,T :C7→C a nonexpansive mapping with F ix(T)6=∅, and f an α-contraction. Then when the sequence {αn} satisfiesH3,1 the sequence {xn} defined by (34)withTn
def=T converges strongly to Q(f).
Proof : Here the sequenceTn does not depend onn. We just apply Corollary 4 to get the result. Of course, if the sequence{xn}is bounded then{Tn(xn) = T xn}is bounded and equation (8) ofH1,1is then satisfied withδn=|αn−αn+1|.
Since{αn} satisfies H3,1, {δn} satisfies H1,1. We also have kTnxn−T xnk =
0→0 and the conclusion follows.
Remark 11 Suppose now that T def=P
iλiTi where {λi} is a sequence of pos- itive real numbers such that P
iλi = 1 and the Ti mappings are all supposed nonexpansive. Then, we can apply Theorem 10 to obtain the strong conver- gence of the sequence {xn} to QF ix T(f). Moreover, If we assume that X is strictly convex then using remark 9 we obtain a strong convergence to QF(f) withF def=∩i∈IF ix(Ti).
This can be extended to the case when theλialso depends onnand recover [9, Theorem 4] as follows :
Corollary 12 LetX be a strictly convex B real Banach space,Ca closed convex subset ofX,Ti:C7→C for i∈I a finite family of nonexpansive mapping with
∩i∈IF ix(Ti)6=∅, andf anα-contraction. For a sequence{αn}satisfyingH3,1 we consider the sequence{xn} defined by (34)with Tn
def=P
i∈Iλi,nTi. Assume that for all i andn λi,n ∈[a, b] with a > 0 andb <∞ either P
nλi,n <∞ or λi,n/αn →0 then{xn} converges strongly toQF(f)with F=∩i∈IF ix(Ti)
Proof :The proof is given by an application of corollary 5. Indeed since the λi,n are bounded Tnxn remains bounded for a bounded sequencexn. ThenTn
satisfiesH′1,1withρn=P
i∈Iλi,n. By extracting from each given subsequence σ(n) a subsequence µ(n) such that limn→∞λi,µ(n) = λi for all i ∈ I we can use corollary 5. Finally, noting that, for a strictly convex space X, the fixed points ofTλdef=P
i∈IλiTidoes not depend onλand is equal to∩i∈IF ix(Ti) we
conclude the proof.
2.2 Example 1
′In [14] The following algorithm is considered :
yn+1=P(αnf(yn) + (1−αn)T yn) (12) Where P : X 7→ C is a sunny nonexpansive retraction, f : C 7→ X an α- contraction andT :C7→X a nonexpansive mapping such thatF ix(T)6=∅.
If we consider the sequence xn+1 =αnf(yn) + (1−αn)T yn then we have yn+1=P xn+1 and thus
xn+1 =αnf(P(xn)) + (1−αn)T(P(xn)) (13) Sincef◦Pis anα-contraction fromXontoXandT◦Pa non expansive mapping fromXontoXwe can use the previous theorem to obtain the strong convergence of the sequence{xn}toxa fixed point ofT◦P such thatx=PF ix(T◦P)f(T(x)) (PS is the metric projection onS). We thus obtain now the strong convergence of the initial sequence{yn}toy=P(x) and sincexis a fixed point ofT◦P,y is a fixed point ofP◦T.
If we suppose in addition that X is such that J (or Jφ) is norm-to-weak⋆ continuous (i.eXis smooth) and thatTsatisfy the weakly inward condition then
we can use the result of [14, Lemma 1.2] which state thatF ix(T) =F ix(P◦T) to conclude thaty is in fact a fixed point of T and recover the result of [14, Theorem 2.4].
2.3 Example 2
We consider now the example given in [8] where the sequence{xn}is given by : yn=βnxn+ (1−βn)T xn
xn+1=αnu+ (1−αn)yn
With a sequence of mappingsTnxdef=βnx+(1−βn)T x. This problem is rewritten as follows :
xn+1=αnf(xn) + (1−αn)Tnxn (14) Theorem 13 Let X be aBreal Banach space, C a closed convex subset ofX, T :C7→C a nonexpansive mapping with F ix(T)6=∅, andf an α-contraction.
When the sequence{αn}satisfiesH3,1and the sequence{βn}converges to zero and satisfy either P∞
n=0|βn+1−βn| <∞ or |βn+1−βn|/αn → 0. Then, the sequence{xn} defined by (14)converges strongly to Q(f).
This theorem is very similar to [8, Theorem 1] wheref was supposed to be constant. It could be covered by corollary 12 but here strict convexity is not needed.
Proof : We easily check that the fixed points p of T are fixed points of Tn for all n ∈ N and Tn is nonexpansive for all n. Thus by Lemma 23 the sequence{xn} is bounded . If the sequence{xn} is bounded thenkTn(xn)k ≤ max(kxnk,kT xnk)}is bounded too. Since :
kTnyn−T ynk ≤βn(kynk+kT ynk) (15) we havekTnyn−T ynk →0 for each bounded sequence{yn}. It is easily checked thatH1,1is satisfied withδn=|αn+1−αn|+|βn+1−βn|. The conclusion follows
from Corollary 4.
2.4 Example 3
We consider here the accretive operators example given in [8] or [18] :
xn+1=αnf(xn) + (1−αn)Tnxn (16) WhereTnx=JrnxandJλis the resolvent of anm-accretive operatorA,Jλx= (I+λA)−1. The following theorem is similar to [18, Theorem 4.2, Theorem 4.4]
or [8, Theorem 2].
Theorem 14 Let X be a B real Banach space, A an m-accretive operator in X such that A−1(0) 6=∅. We assume here that C def=D(A) whereD(A) is the domain of A and suppose that C is convex. Suppose that H3,1 is satisfied by the sequence {αn} and that the sequencern is such thatrn ≥ǫ >0 and either P∞
0 |1−rn/rn+1|<∞or|1−rn/rn+1|/αn→0, then the sequence{xn}defined by (16)converges strongly to a zero of A.
Proof : We first note that [18, p 632], forλ >0, F ix(Jλ) =F where F is the set of zero ofAand for anm-accretive operatorA,Jλis non expansive from X 7→D(A). Using the resolvent identityJλx=Jµ((µ/λ)x+ (1−µ/λ)Jλx) we obtain :
kTn+1zn−Tnznk ≤ 1− rn
rn+1
(kznk+kTnznk) (17) and since the sequence Tnyn is bounded for a bounded sequence yn (for p ∈ A−1(0) we have kTnyn−pk ≤ kyn−pk) we can apply remark 2 in order to obtainH1,1,.We thus havekxn+1−xnk →0 by Lemma 24 andkxn−Tnxnk → 0 by :
kxn−Tnxnk ≤ kxn−xn+1k+kxn+1−Tnxnk
≤ kxn−xn+1k+αn(kf(xn)k+kTn(xn)k) Take nowrsuch that 0< r < ǫand defineT def=Jrthen we have :
kTnxn−T xnk ≤ 1− r
rn
kxn−Tnxnk (18) We thus obtain thatxn−T xn→0 from :
kxn−T xnk ≤ kxn−Tnxnk+kTnxn−T xnk (19)
The conclusion is obtained through Corollary 4.
2.5 Example 4
We consider here the example given in [13]
xn+1=αnf(xn) + (1−αn)Tnyn (20) whereTn=QnmodN, whereN ≥1 is a fixed integer and the (Ql)l=0,...,N−1 is a family of nonexpansive mappings.
Theorem 15 Let X be aBreal Banach space, C a closed convex subset ofX, Ql : C 7→ C for l ∈ {1, . . . , N} a family of nonexpansive mappings such that F def=∩Nl=0−1F ix(Ql)is not empty and
∩Nl=0−1F ix(Ql) =F ix(Tn+NTn+N−1· · ·Tn+1)for alln∈N (21) and f an α-contraction. When the sequence {αn} satisfies H3,N then the se- quence{xn} defined by (20)converges strongly to QF(f).
Proof : By Lemma 23, since theTnhave a common fixed point, the sequence {xn} is bounded. Since the sequence of mappingsTn is periodic, the sequence {Tnxn}is bounded and equation (8) ofH1,N is obtained forδn=|αn−αn+N| using (9). Since{αn} satisfiesH3,N, {δn} satisfiesH1,N. Thus, using Lemma 24 we obtain that kxn+N −xnk → 0. Since kxn+1−Tnxnk ≤ αn(kf(xn)k+ kTnxnk), we havekxn+1−Tnxnk →0. We introduce the sequence of mappings A(N,α)n
def= Tn+N−1· · ·Tn+α for α 6= N and A(N,N)n = Id. Using Lemma 16, given just after this proof, we conclude that : kxn+N −A(N,0)n xnk → 0. This combined with kxn+N−xnk → 0 gives kxn+N −A(N,0)n xnk → 0. Note now that the mappingsA(N,0)n are in finite number are all nonexpansive and share common fixed points by hypothesis. Thus we can prove thatH2,p is satisfied forp=QF(f). Let p=QF(f) we suppose that H2,p is not satisfied, then it possible to extract a subsequence of{xσ(n)}such that :
n→∞lim
f(p)−p, J(xσ(n)−p)
≤0 (22)
But it is then possible to findq∈ {0, . . . , N−1}and an extracted new subse- quenceµ(n) fromσ(n) such thatµ(n)modN =q. We thus havekxµ(n)−T xµ(n)k → 0, withT def=A(N,0)q which is now a fixed mapping andF ix(T) =F. ThenH2p should be true by Lemma 26 and this leads to a contradiction. The conclusion
follows by 28.
Lemma 16 LetN ∈N,α∈ {0, . . . , N}andA(N,α)n
=defTn+N−1· · ·Tn+αforα6=
NandA(N,Nn )=Id. Assume thatkxn+1−Tnxnk →0thenkxn+N −A(N,0)n xnk → 0.
Proof : We have for α∈ {0, . . . , N −1} by definition of A(N,α)n and using the fact thatA(N,α)n is nonexpansive :
kA(N,α+1)n xn+α+1−A(N,α)n xn+αk = kA(N,α+1)n xn+α+1−A(N,α+1)n Tn+αxn+αk
≤ kxn+α+1−Tn+αxn+αk Thus :
kxn+N−A(N,0)n xnk ≤
N−1X
α=0
kxn+α+1−Tn+αxn+αk
and the result follows.
2.6 Example 5
Let Γ(j)n for j ∈ {1, . . . , m} be a sequence of mappings defined recursively as follows :
Γ(j)n xdef=β(j)n x+ (1−βn(j))TjΓ(j+1)n xand Γ(m+1)n x=x (23)
where the sequences{βn(j)} ∈(0,1), and {Tj} forj∈ {1, . . . , m} are nonexpan- sive mappings. We want to prove here the convergence of the sequence generated by the iterations :
xn+1=αnf(xn) + (1−αn)Γ(1)n xn (24) Theorem 17 Let X be aBreal Banach space, C a closed convex subset ofX, Tj : C 7→ C for j ∈ {1, . . . , m} a family of nonexpansive mappings such that
∩ml=1F ix(Tj) is not empty and f an α-contraction. When the sequence {αn} satisfiesH3,Nand forj∈ {1, . . . , m}the sequences{βn(j)}satisfylimn→∞βn(j)= 0and eitherP∞
n=0|βn+1(j) −βn(j)|<∞or|β(j)n+1−βn(j)|/αn→0then the sequence defined by (24)converges strongly toQF(f)associated toF =F ix(T1· · ·Tm).
Proof : Note first that by an elementary induction Γ(1)n is a nonexpansive mapping. If we assume thatpis a common fixed point to the mappingsTithen p is a fixed point of the mappings Γ(j)n . By Lemma 23 the sequence {xn} is bounded. Then using Lemma 19 , given just after this proof, combined with the boundedness of{xn},H1,1is valid with
δn= Xm p=1
|βn+1(p) −βn(p)|+|αn+1−αn| (25) Now if we can prove that
kΓ(1)n xn−T1T2· · ·Tmxnk →0 (26) the conclusion will be given by Corollary 4. The last assetion can easily be obtained by induction onkΓ(j)n xn−Tj· · ·Tmxnk, since we have :
kΓ(j)n xn−Tj· · ·Tmxnk ≤ β(j)n (kxnk+kTj· · ·Tmxnk)
+(1−βn)kTjΓ(j+1)n xn−Tj· · ·Tmxnk
≤ β(j)n (kxnk+kTj· · ·Tmxnk) +kΓ(j+1)n xn−Tj+1· · ·Tmxnk.
Remark 18 Form= 1we obtain the same result as Theorem 13.
Lemma 19 LetΓ(j)n be the sequence of mappings defined by (23) Then we have forj ∈ {1, . . . , m} :
kΓ(j)n+1x−Γ(j)n xk ≤
Xm p=j
|βn+1(p) −βn(p)|
K (27) whereK is a constant which depends on the mappings (Tp)p≥j andx.
Proof : Note first that :
kΓ(j)n xk ≤ kxk+kTj(Γ(j+1)n x)k (28) which applied recursively shows that kΓ(j)n xk is bounded by a constant which depends on the mappings (Tp)p≥jandxand not onn. Then, using the definition of Γ(j)n we have :
kΓ(j)n+1x−Γ(j)n k ≤ |βn+1(j) −βn(j)|(kxk+kTjΓ(j+1)xk)
+kTjΓ(j+1)n+1 (x)−TjΓ(j+1)n (x)k (29) sinceTj is nonexpansive mappings :
kΓ(j)n+1x−Γ(j)n k ≤ |βn+1(j) −βn(j)|(kxk+kTjΓ(j+1)xk) +kΓ(j+1)n+1 (x)−Γ(j+1)n (x)k by recursion and since the last term Γ(m+1)n+1 (x)−Γ(m+1)n (x) = 0 we obtain the
result.
Note that Lemma 19 remains valid for the sequence
Γ(j)n xdef=βn(j)g(x) + (1−βn(j))TjΓ(j+1)n xand Γ(m+1)n x=x (30) ifg is a nonexpansive mapping.
2.7 Example 6
We consider here the example given in [3]
xn+1=αnf(xn) + (1−αn)Tnxn
where Tnx def= PC(x−λnAx) and PC is the metric projection from X to C.
The aim is to find a solution of the variational inequality problem which is to findx∈C such thathAx, y−xi ≥0 for all y ∈C. The set of solution of the variational inequality problem is denoted by VI(C, A). The operatorA is said to beµ-inverse-strongly monotone if
hx−y, Ax−Ayi ≥µkAx−Ayk2 for allx, y ∈C The next theorem is similar to [3, Proposition 3.1].
Theorem 20 Let X be a real Hilbert space, C a nonempty closed convex, f an α-contraction, and let A be a µ-inverse-strongly monotone mapping of H into itself such that VI(C, A) 6= ∅. Assume that H3,1 is satisfied and that {λn} is chosen so that λn ∈ [a, b] for some a, b with 0 < a < b < 2µ and P∞
n=1|λn+1−λn| < ∞. then the sequence {xn} generated by (31) converges strongly to QF(f) associated to F = F ix(Tλ) where Tλ(x) def= PC(x−λAx).
F =F ix(Tλ) does not depend onλfor λ >0 and equals VI(C, A).
Proof :Forλ >0, letTλxdef= PC(x−λAx). When X is an Hilbert space we have F ix(Tλ) = VI(C, A). When A is µ-inverse-strongly monotone then for, λ ≤ 2µ, I−λA is nonexpansive. Thus the mappings Tn are non expansive and F ix(Tn) = VI(C, A)6= ∅. By Lemma 23 the sequence{xn} is bounded.
Since kTnzk ≤ K(kzk+ 2µkAzk), the sequence {Tnxn} is bounded too. We also havekTn+1zn−Tnznk ≤ |λn+1−λn|kAznk which gives H1,N with δn =
|λn+1−λn|+|αn+1−αn| by remark 2. The result follows now from Corollary 5. Indeed, sinceλσ(n)∈[a, b] it is possible to extract a converging subsequence λµ(n)→ λ∈[a, b] and we then havekTµ(n)z−Tλzk ≤ |λµ(n)−λ|kAzk. Thus
kTµ(n)xµ(n)−Tλxµ(n)k →0.
Remark 21 We can note that forλ <2α,I−λA is in fact attracting nonex- pansive since :
k(I−λA)x−(I−λA)yk ≤ kx−yk+λ(λ−2α)kAx−Ayk2.
Thus it is also the case for PC◦(I−λA)[1]. For a nonexpansive mapping S we can consider the previous theorem withTλxdef=S◦PC(x−λAx) and using Remark 7 (an Hilbert space is strcitly convex) to obtain a strong convergence to a point in F ix(Tλ) = F ix S∩VI(C, A) and thus fully recover [3, Proposition 3.1]
2.8 Example 7
We consider here the equilibrium problem for a bifunctionF :C×C7→Rwhere C is a closed convex subset of a real Hilbert space X. The problem is to find x∈ C such thatF(x, y)≥0 for ally ∈C. The set of solutions if denoted by EP(F). It is proved in [5] (See also [4]) that forr >0, the mappingTr:X7→C defined as follows :
Tr(x)def=
z∈C:F(z, y) + 1
rhy−z, z−xi ≥0,∀y∈C
(31) is such that Tr is singled valued, firmly nonexpansive (i.e kTrx−Tryk2 ≤ hTrx−Try, x−yifor anyx, y∈X),F ix(Tr) = EP(F) and EP(F) is closed and convex if the bifunctionF satisfies (A1)F(x, x) = 0 for allx∈C, (A2)F(x, y) + F(y, x)≤0 for allx,y∈C, (A3) for eachx,y,z∈Climt→0F(tz+(1−t)x, y)≤ F(x, y) and (A4) for eachx∈C y7→F(x, y) is convex and lower semicontinu- ous.
we can now consider the sequence{xn}given by : xn+1=αnf(xn) + (1−αn)Tnxn
whereTn
def=Trn for a given sequence of real numbers{rn}.
Theorem 22 Let X be a real Hilbert space,C a nonempty closed convex,f an α-contraction, assume that EP(F)6=∅,H3,1is satisfied and the sequence{rn}is such thatlim infn→∞rn>0and eitherP
n|rn+1−rn|<∞of|rn+1−rn|/αn→ 0. Then, the sequence{xn}generated by (32)converges strongly toQEP(F)(f).
Proof : Since thern are strictly positive the mappingsTrn are non expansive and share the same fixed points EP(F) which was supposed non empty. By Lemma 23 the sequence{xn}is bounded.
Using the definition of Tr(x) and the monotonicity ofF (A2) easy compu- tations leads to the following inequality [12, p 464] :
kTr(x)−Ts(y)k ≤ kx−yk+1−s r
kTr(y)−yk (32) Using r > 0 such that rn > r for all n ∈ N and y ∈ F ix(Tr) we obtain kTrn(xn)−Tr(y)k ≤ kxn−yk which gives the boundedness of the sequence {Trn(xn)}. Moreover, for a bounded sequence{yn} we obtain :
kTrn+1(yn)−Trn(yn)k ≤ |rn+1−rn|
r kTrn(yn)−ynk (33) We thus obtain H1,1 with δn = |rn+1 −rn|+|αn+1 −αn| using remark 2.
The result follows now from Corollary 5. Indeed, since rσ(n) > r it is pos- sible to extract a converging subsequence rµ(n) → r > r and we then have kTrµ(n)z−Trzk ≤ |rµ(n)−r|K. Thus
kTrµ(n)xµ(n)−Trxµ(n)k →0.
3 A collection of Lemma
The first Lemma can be used to derive boundedness of the sequence{xn} gen- erated by 34.
Lemma 23 Let {xn}, the sequence generated by the iterations
xn+1=αnf(xn) + (1−αn)Tnxn (34) wheref is contraction of parameterα,Tn is a family of nonexpansive mappings andαn is a sequence in(0,1). Suppose that there existspa common fixed point ofTn for all n∈N. Then, the sequence{xn} is bounded.
Proof : The proof exactly follows the proof of [17, theorem 3.2], the only differ- ence is that here the mappingsTn are indexed bynbut it does not change the
proof. Obviously we have :
kxn+1−pk ≤ αnkf(xn)−pk+ (1−αn)kTnxn−pk
≤ αn(αkxn−pk+kf(p)−pk) + (1−αn)kxn−pk
≤ (1−αn(1−α))kxn−pk+αn(1−α)kf(p)−pk (1−α)
≤ max
kxn−pk,kf(p)−pk (1−α)
.
And, by induction,{xn} is bounded.
The next lemma aims at proving that the sequence {xn} is asymptotically regulari.efor a givenN≥1, we havekxn+N −xnk →0.
Lemma 24 With the same assumptions as in Lemma 23 and assuming that there existsN ≥1 such thatH1,N andH3,N are fulfilled then, for the sequence {xn} given by iterations (34), we havekxn+N −xnk →0.
Proof : Using the definition of{xn} we have :
xn+N+1−xn+1 = αn+N(f(xn+N)−f(xn)) + (αn+N−αn)f(xn) +(1−αn+N)(Tn+Nxn+N −Tn+Nxn)
+ ((1−αn+N)Tn+Nxn−(1−αn)Tnxn).
By Lemma 23 the sequence{xn} is bounded, we can therefore use H1,N with {xn}. Since{f(xn)}is bounded too, we can find three constants such that : kxn+N+1−xn+1k ≤ αn+Nαkxn+N −xnk+|αn+N−αn|K1
+(1−αn+N)kxn+N−xnk+δnM
≤ (1−(1−α)αn+N)kxn+N −xnk+ (|αn+N −αn|+δn)K2
The proof then follows easily using the properties ofαni.eH3,Nand Lemma 30.
The next step is to prove that we can find a fixed mapping T such that kxn−T xnk → 0. The next corollary gives a simple example for which the property can be derived from Lemma 24. Indeed, we have seen specific proofs in previous sections on illustrated examples.
Corollary 25 Using the same hypothesis as in Lemma 24 and assuming that {Tnxn}is bounded and thatkTnxn−T xnk →0 we also have kxn−T xnk →0.
Proof :
kxn−T xnk ≤ kxn−xn+1k+kxn+1−T xnk
≤ kxn−xn+1k+αnK1+ (1−αn)kTnxn−T xnk
and the result follows.
The next Lemma gives assumptions to obtainH2,pfor a givenp.
Lemma 26 Suppose thatX is aBreal Banach space. LetT be a nonexpansive mapping with F ix(T) 6= ∅, f an α-contraction and {xn} a bounded sequence such thatkT xn−xnk →0. Then forx˜=Q(f)we have :
lim sup
n→∞
hf(˜x)−˜x, J(xn−x)i ≤˜ 0 (35)
Proof :WhenX is aBus or aBrug the key point is the fact thatJ is uniformly norm-to-weak⋆continuous on bounded sets.
The proof of this Lemma can be found in the proof of Theorem [17, Theorem 4.2] or [13, Theorem 3.1]. We just summarize the line of the proof here. Let
˜
xdef=σ- limt→0xt wherext solvesxt=tf(xt) + (1−t)T xt, we thus have : kxt−xnk2 ≤ (1−t)2kT xt−xnk2+ 2thf(xt)−xn, J(xt−xn)i
≤ (1−t)2(kT xt−T xnk+kT xn−xnk)2 +2thf(xt)−xt, J(xt−xn)i+ 2tkxt−xnk2
≤ (1 +t2)kxt−xnk2+an(t) +2thf(xt)−xt, J(xt−xn)i
(36) where an(t) = 2kT xn−xnkkxt−xnk+kT xn−xnk2 → 0 when n tends to infinity. Thus :
hf(xt)−xt, J(xn−xt)i ≤ an(t) 2t + t
2kxt−xnk2 (37) and we have :
t→0limlim sup
n→∞
hf(xt)−xt, J(xn−xt)i ≤0 (38) We consider now a sequencetp → 0 and yp
def= xtp, then we have yp → x˜ and withg(x)def= (x)−xwe have
hg(˜x), J(xn−x)i ≤ hg(y˜ p), J(xn−yp)i
+ | hg(˜x), J(xn−x)˜ −J(xn−yp)i |+ (1 +α)k˜x−ypkkxn−ypk SinceJ is uniformly norm-to-weak⋆continuous on bounded sets andyp→x,˜ forǫ >0, we can find ˜psuch that for allp≥p˜and alln∈Nwe have :
hg(˜x), J(xn−x)i ≤ hg(y˜ p), J(xn−yp)i+ǫ(1 +α)k˜x−ypkkxn−ypk(39)
Thus : lim sup
n→∞
hg(˜x), J(xn−x)i ≤˜ lim sup
n→∞
hg(yp), J(xn−yp)i+ǫ+k˜x−ypkK
≤ lim
p→∞(lim sup
n→∞
hg(yp), J(xn−yp)i+ǫkx˜−ypkK)≤ǫ Suppose now thatX is aBrwsc. We follow the proof of [Theorem 2.2]song- chen-1 or [18, Theorem 3.1]. Let ˜x=Q(f) and consider a subsequence{xσ(n)} such that lim supn→∞hf(˜x)−x, J(x˜ n−x)i˜ = limn→∞
f(˜x)−x, J˜ (xσ(n)−x)˜ . It is then possible to re-extract a subsequence xµ(n) weakly converging to x⋆. Since we have xµ(n)−T xµ(n) → 0 then x⋆ ∈ F ix(T) using the key property that X satisfies Opial’s condition [7, Theorem 1] and the fact that I −T is demi-closed at zero [13, Lemma 2.2]. Thus by definition of ˜x we must have
hf(˜x)−x, J(x˜ ⋆−x)i ≤˜ 0.
Corollary 27 Suppose that X is a Bus, or a Brug, or a Brwsc. let f a con- traction and {xn} a bounded sequence such that xn−Tnxn → 0. From each subsequenceσ(n) we can extract a subsequence µ(n) and find a fixed mapping Tµsuch thatkTµ(n)xµ(n)−Tµxµ(n)k →0. Then, ifF=F ix Tµdoes not depend onµ, for x˜=Q(f)associated toF, we have :
lim sup
n→∞
hf(˜x)−˜x, J(xn−x)i ≤˜ 0 (40)
Proof :The proof is by contradiction using Lemma 26. Assume that the result is false, then we can find a subsequenceσ(n) such that
lim sup
n→∞
f(˜x)−x, J˜ (xµ(n)−˜x)
≥ǫ >0 (41)
by hypothesis we can extract fromσ(n) a sub-sequenceµ(n) such thatkTµ(n)xµ(n)−T xµ(n)k → 0. Thus, since
kxµ(n)−T xµ(n)k ≤ kxµ(n)−Tµ(n)xµ(n)k+kTµ(n)xµ(n)−T xµ(n)k, we have xµ(n)−T xµ(n) → 0 we can then apply Lemma 26 to the sequence {xµ(n)} and mappingTµ to derive that :
lim sup
n→∞
f(˜x)−x, J˜ (xµ(n)−x)˜
≤0
for ˜x=Q(f) corresponding to F =F ix Tµ and sinceF does not depend onµ,
this gives a contradiction with (41).
The next Lemma helps concluding the proof.
Lemma 28 Assume that the sequence{xn}given by iterations (34) is bounded and assume that forp, a common fixed point of the mappingsTn,H2,pis satisfied and that (i, ii, iii) items of H3,N is also satisfied1. Then the sequence {xn} converges top.
Proof :
kxn+1−pk2 ≤ (1−αn)2kTnxn−pk2+ 2αnhf(xn)−p, J(xn+1−p)i
≤ (1−αn)2kxn−pk2+ 2αnhf(xn)−f(p), J(xn+1−p)i +2αnhf(p)−p, J(xn+1−p)i
≤ (1−αn)2kxn−pk2+ 2αnαkxn−pkkxn+1−p)k +2αnhf(p)−p, J(xn+1−p)i
Note thatkxn+1−pk ≤ kxn−pk+αnK . Thus :
kxn+1−pk2 ≤ (1−αn)2kxn−pk2+ 2αnαkxn−pk2 +2α2nK+ 2αnhf(p)−p, J(xn+1−p)i
≤ (1−αn(1−α) +α2n)kxn−pk2 +2α2nK+ 2αnhf(p)−p, J(xn+1−p)i
(42)
And we conclude with Lemma 29.
Lemma 29 .[8, Lemma 2.1] Let{sn}be a sequence of nonnegative real numbers satisfying the property
sn+1≤(1−αn)sn+αnβn for n≥0,
whereαn∈(0,1)andβnare sequences of real numbers such that : (i) limn→∞αn= 0andP∞
n=0αn =∞(ii)eitherlim supn→∞βn≤0orP∞
n=0|αnβn|<∞. Then {sn} converges to zero.
Corollary 30 Let {sn} be a sequence of nonnegative real numbers satisfying the property
sn+1≤(1−αn)sn+αnβn+αnγn for n≥0,
where αn ∈ (0,1), βn and γn are sequences of real numbers such that : (i) limn→∞αn = 0andP∞
n=0αn =∞(ii) lim supn→∞βn ≤0and(iv)P∞
n=0|αnδn|<
∞. Then{sn}converges to zero.
Proof :The proof is similar to the proof of Lemma 29 [8, Lemma 2.1]. Fix ǫ > 0 andN such that βn ≤ ǫ/2 for n ≥ N and P∞
j=N|αnδn| ≤ ǫ/2 . Then
1Note that (i, ii, iii) ofH3,Ndo not use the value ofN
following [8] we have forn > N : sn+1 ≤
Yn j=N
(1−αj)sN +ǫ 2(1−
Yn j=N
(1−αj)) + Xn j=N
|αnδn|
≤ Yn j=N
(1−αj)sN +ǫ 2(1−
Yn j=N
(1−αj)) + ǫ
2 (43)
and then by taking the limit sup whenn→ ∞we obtain lim supn→∞sn+1≤ǫ.
A contraction is said to be a Meir-Keeler contraction (MKC) if for every ǫ >0 there exitsδ >0 such thatkx−yk< ǫ+δimplieskΦ(x)−Φ(y)k< ǫ.
Lemma 31 [15] Suppose that the sequence{xn}defined by equation (34) strongly converges for an α-contraction f (or a constant function f) to the fixed point ofPF◦f then the results remains valid for a Meir-Keeler contraction Φ.
Proof :Suppose that we have proved that (34) converges for anα-contraction f to the fixed point of PF ◦f. Then indeed, the result is true when f is a constant mapping. Let Φ be a Meir-Keeler contraction, fix y ∈C, when f is constant and equal to Φ(y) then{xn} defined by (34) converges to PF(Φ(y)).
If Φ is a MKC then sincePF is nonexpansivePF◦Φ is also MKC (Proposition 3 of [15]) and has a unique fixed point [10]. We can considerz=PF(Φ(z)) and consider two sequences :
xn+1=αnΦ(xn) + (1−αn)Tnxn (44)
yn+1=αnΦ(z) + (1−αn)Tnyn (45) Of course{yn}converges strongly toz. We now prove that{xn}also converges strongly to z following [15]. Fix ǫ > 0, by Proposition 2 of [15], we can find r∈(0,1) such thatkx−yk ≤ǫimplieskΦ(x)−Φ(y)k ≤rkx−yk. Choose now N such that kyn−zk ≤ ǫ(1−r)/r. Assume now that for all n≥ N we have kxn−ynk> ǫthen
kxn+1−yn+1k ≤ (1−αn)kxn−ynk+αnkΦ(xn)−Φ(yn)k+αnkΦ(yn)−zk
≤ (1−αn(1−r))kxn−ynk+αnǫ
We cannot use here directly Lemma 29 but following the proof of this Lemma we obtain that lim supkxn−ynk ≤ ǫ. Assume now that for a given value of n we have kxn−ynk ≤ ǫ. Since Φ is a MKC we have kΦ(x)−Φ(y)k ≤ max(rkx−yk, ǫ) and since we have
rkxn−zk ≤rkxn−ynk+rkyn−zk ≤ǫ (46)
we obtain
kxn+1−yn+1k ≤(1−αn)kTnxn−Tnynk+αnmax(rkxn−zk, ǫ)≤ǫ . (47) Thus we have in both cases lim supn→∞kxn−ynk ≤ǫand the conclusion fol-
lows.
Lemma 32 [1, Proposition 2.10 (i)] Suppose that X is strictly convex, T1 an attracting non expansive mapping andT2 a non expansive mapping which have a common fixed point. Then :
F ix(T1◦T2) =F ix(T2◦T1) =F ix(T2)∩F ix(T1).
Proof :We haveF ix(T2)∩F ix(T1)⊂F ix(T2◦T1) andF ix(T2)∩F ix(T1)⊂ F ix(T1◦T2). Let xbe a common fixed point ofT1 andT2. Ify, a fixed point ofT1◦T2, is such that y 6∈F ix(T2) then sinceT1 is attracting non expansive we have :
ky−xk=kT1◦T2(y)−xk<kT2(y)−xk ≤ ky−xk
which gives a contradiction. Thusy is a fixed point ofT2 and then also of T1. If nowy a fixed point ofT2◦T1 and assume thaty6∈F ix(T1) then we have
ky−xk=kT2◦T1(y)−xk ≤ kT1(y)−xk<ky−xk
which gives also a contradiction and same conclusion.
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